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Stoichiometric Calculations

The ionizatio...

Question

The ionization constant of dimethylamine is $5.4 \times 10^{- 4}$ . Calculate its degree of ionization in its 0.02 M solution. What percentage of dimethylamine is ionized if the solution is also 0.1 M in NaOH?

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Solution

$K_{b} = 5.4 \times 10^{- 4} c = 0.02 M$
Then $α = \sqrt{\frac{K_{b}}{c}} = \sqrt{\frac{5.4 \times 10^{- 4}}{0.02}} = 0.1643$
Now, if 0.1 M of NaOH is added to the solution, then NaOH (being a strong base) undergoes complete ionization.

$\begin{matrix} N a O H_{(a q)} & \leftrightarrow & N a_{(a q)}^{+} & + & O H_{a q}^{-} 0.1 M & 0.1 M \end{matrix}$
And,
$\begin{matrix} (C H_{3})_{2} N H & + & H_{2} O & \leftrightarrow & (C H_{3})_{2} N H_{2}^{+} & + & O H (0.02 - x) & x & x 0.02 M & 0.1 M \end{matrix}$
Then, $[(C H_{3})_{2} N H_{2}^{+}] = X [O H^{-}] = x + 0.1 \Rightarrow K_{b} = \frac{[(C H_{3})_{2} N H_{2}^{+}] [O H^{-}]}{[(C H_{3})_{2} N H]} 5.4 \times 10^{- 4} = \frac{x \times 0.1}{0.02} x = 0.0054$
It means that in the presence of 0.1 M NaOH, 0.54% of dimethylamine will get dissociated.

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The ionization constant of dimethylamine is 5.4×10−4. Calculate its degree of ionization in its 0.02 M solution. What percentage of dimethylamine is ionized if the solution is also 0.1 M in NaOH?

The ionization constant of dimethylamine is $5.4 \times 10^{- 4}$ . Calculate its degree of ionization in its 0.02 M solution. What percentage of dimethylamine is ionized if the solution is also 0.1 M in NaOH?